Integrating e^x & 1/x

$\int e^{x} d x = e^{x} + c$

$\int \frac{1}{x} d x = \ln | x | + c$

where $c$ is the constant of integration

$\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$

$\int \frac{1}{a x + b} d x = \frac{1}{a} \ln | a x + b | + c$

$\int \frac{a}{a x + b} d x = \ln | a x + b | + c$

- which can be deduced using Reverse Chain Rule
With ln, it can be useful to write the constant of integration, $c$ , as a logarithm
- using the laws of logarithms, the answer can be written as a single term
- $\int \frac{1}{x} d x = \ln | x | + \ln k = \ln k | x |$ where $k$ is a constant
- This is similar to the special case of differentiating $\ln (a x + b)$ when $b = 0$

A curve has the gradient function $f' (x) = \frac{3}{3 x + 2} + e^{4 - x}$ .

Given the exact value of $f (1)$ is $\ln 10 - e^{3}$ find an expression for $f (x)$ .

To find $f (x)$ , we need to integrate the expression for $f' (x) .$

$f (x) = \int \frac{3}{3 x + 2} + e^{4 - x} d x$

Rewrite as two separate integrals which can be found individually, and factor out any constants.

$f (x) = 3 \int \frac{1}{3 x + 2} d x + \int e^{4 - x} d x$

Use the two results:
$\int \frac{1}{a x + b} d x = \frac{1}{a} \ln | a x + b | + c$
$\int e^{a x + b} d x = \frac{1}{a} e^{a x + b} + c$

$f (x) = 3 (\frac{1}{3} \ln | 3 x + 2 |) + - 1 e^{4 - x} + c f (x) = \ln | 3 x + 2 | - e^{4 - x} + c$

The question states that $f (1) = \ln 10 - e^{3}$ , so substitute in $x = 1$ and equate to the given expression.

$\ln 10 - e^{3} = \ln | 3 (1) + 2 | - e^{4 - 1} + c$

Simplify and solve to find $c .$

$\ln 10 - e^{3} = \ln 5 - e^{3} + c \ln 10 - \ln 5 = c$

Recall from laws of logarithms that $\ln a - \ln b = \ln \frac{a}{b} .$

$c = \ln \frac{10}{5} = \ln 2$

Rewrite the full expression for $f (x) .$

$f (x) = \ln | 3 x + 2 | - e^{4 - x} + \ln 2$
or by combining logs, this could be written as $f (x) = \ln (2 | 3 x + 2 |) - e^{4 - x}$

Integrating e^x & 1/x (Cambridge O Level Additional Maths)